A race car moves such that its position ?ts the relationship x(t) = (5.3 m/s) t
ID: 2025370 • Letter: A
Question
A race car moves such that its position ?tsthe relationship
x(t) = (5.3 m/s) t + (0.77 m/s^3) t^3
where x is measured in meters and t in seconds.
A.) Approximate the instantaneous velocity at
1.1 s, using a centered time interval of 0.3 s.
Answer in units of m/s.
B.) Use a 0.6 s interval centered at 1.1 s to approximate the instantaneous velocity at 1.1 s.
Answer in units of m/s.
C.) Find the average velocity during the 1.1 s.
Answer in units of m/s.
I am having trouble understanding what I have to do here. Can someone help me solve this problem and help me understand for my test? Thanks in advance.
Explanation / Answer
Instantaneous velocity is just the derivative of the X displacement with respect to time. Remember that instantaneous velocity=dx/dt....which means that velocity is the derivative of displacement with respect to time.
So we must take the derivative of x(t) to get our equation for v(t):
x(t)= (5.3m/s)t + (0.77m/s)t^3
V(t)= 5.3 + (3)(0.77)t^2
=5.3 +(2.31)t^2
the time you would plug in would be the place on the graph that you are approximating. I would ask your professor what they mean by centered means because the question is somewhat ambiguous....you should be able to plug in the time for that particular instant of time that you want to find the velocity at but I may be wrong...there are several ways of approximating solutions..but taking the derivative or the limit is the most logical.
Average velocity is = (x2-x1)/(t2-t1)
This is different from instantaneous because we first need to find the displacement X1 at t1 and then X2 at t2....then we subtract those displacements that we found...so X2-X1.....then we divide by the time interval (t2-t1)